-2u^2+6u+16=0

Solution for -2u^2+6u+16=0 equation:

-2u^2+6u+16=0

a = -2; b = 6; c = +16;
Δ = b2-4ac
Δ = 62-4·(-2)·16
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{41}}{2*-2}=\frac{-6-2\sqrt{41}}{-4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{41}}{2*-2}=\frac{-6+2\sqrt{41}}{-4}
$